Making Sequences is Fun373B

We'll define S(n) for positive integer n as follows: the number of the n's digits in the decimal base. For example, S(893) = 3,S(114514) = 6.

You want to make a consecutive integer sequence starting from number m (m, m + 1, ...). But you need to pay S(n)·k to add the numbern to the sequence.

You can spend a cost up to w, and you want to make the sequence as long as possible. Write a program that tells sequence's maximum length.

Input

The first line contains three integers w (1 ≤ w ≤ 10¹⁶), m (1 ≤ m ≤ 10¹⁶), k (1 ≤ k ≤ 10⁹).

Please, do not write the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64dspecifier.

Output

The first line should contain a single integer — the answer to the problem.

Sample Input

Input

9 1 1

Output

9

Input

77 7 7

Output

7

Input

114 5 14

Output

6

Input

1 1 2

Output

0

#include <stdio.h>
#include <string.h>
long long SS(long long a)
{
    long long x=0;
    while(a>0)
    {
        a=a/10;
        x++;
        //printf("%I64d
",a);
    }
    return x;
}

long long num(long long x)
{
    long long i,y=1;
    for(i=1;i<=x;i++)
        y=y*10;
    return y;
}
int main()
{
    long long w,m,k;
    int i,j;
    while(scanf("%I64d %I64d %I64d",&w,&m,&k)!=EOF)
    {
        long long n=0,nu,ww,mm;
        long long oo=SS(m);
        nu=num(oo);
        ww=w/k;
        while(ww>0)
        {
            mm=ww/oo;
            n=n+mm;
            if(n>=nu-m)
            {
                mm=mm-(n-nu+m);
                n=nu-m;
                ww=ww-oo*mm;
                oo++;
                nu=num(oo);
            }
            else
            {
                ww=ww-oo*mm;
            }
            if(mm==0)
                break;
            //printf("%I64d %I64d %I64d",ww,);
        }
        printf("%I64d
",n);
    }
    return 0;
}